Problem: Simplify the following expression: $r = \dfrac{8y^2 - 8y - 96}{y - 4} $
Solution: First factor the polynomial in the numerator. We notice that all the terms in the numerator have a common factor of $8$ , so we can rewrite the expression: $ r =\dfrac{8(y^2 - 1y - 12)}{y - 4} $ Then we factor the remaining polynomial: $y^2 {-1}y {-12} $ ${-4} + {3} = {-1}$ ${-4} \times {3} = {-12}$ $ (y {-4}) (y + {3}) $ This gives us a factored expression: $\dfrac{8(y {-4}) (y + {3})}{y - 4}$ We can divide the numerator and denominator by $(y + 4)$ on condition that $y \neq 4$ Therefore $r = 8(y + 3); y \neq 4$